TITLE: How to prove $E(X+Y)=E(X)+E(Y)$ using the definition of expectation $E(X)$?
QUESTION [9 upvotes]: Please someone help me to prove the following theorem:
$$\mathbb{E}(X+Y)=\mathbb{E}(X)+\mathbb{E}(Y)$$
using the definition of expectation $\mathbb{E}(X)$
My proof:
$$\mathbb{E}(X+Y) = \sum_{x} \sum_{y}(x+y)P_{XY}(x,y)$$
$$= \sum_{x} \sum_{y}xP_{XY}(x,y) + \sum_{x} \sum_{y}yP_{XY}(x,y)$$
$$= \sum_{x} xP_{X}(x) + \sum_{y}yP_{Y}(y)$$
$$=\mathbb{E}(X)+\mathbb{E}(Y)$$
Is this proof right?

REPLY [3 votes]: Proposition:
Let $X\colon\Omega\to\mathbb{R}^k$ and $Y\colon\Omega\to\mathbb{R}^k$ be random variables in some probability space $(\Omega,\mathcal{F},\mathbb{P})$ and let $\alpha$, $\beta\in\mathbb{R}$. Then,
$$E(\alpha X+\beta Y)=\alpha E(X)+\beta E(Y).$$
Proof:
$$E(\alpha X+\beta Y)=\int_{\Omega}(\alpha X +\beta Y) d\mathbb{P}=\alpha\int_{\Omega}X d\mathbb{P} + \beta\int_{\Omega}Y d\mathbb{P}=\alpha E(X)+\beta E(Y).$$